异步

const fs = require('fs')
const path = require('path')

const readdir = (p, done, a = [], i = 0) => fs.readdir(p, (e, d = []) =>
  d.map(f => readdir(a[a.push(path.join(p, f)) - 1], () =>
    ++i == d.length && done(a), a)).length || done(a))

readdir(__dirname, console.log)

Sync

const fs = require('fs')
const path = require('path')

const readdirSync = (p, a = []) => {
  if (fs.statSync(p).isDirectory())
    fs.readdirSync(p).map(f => readdirSync(a[a.push(path.join(p, f)) - 1], a))
  return a
}

console.log(readdirSync(__dirname))

异步读

function readdir (currentPath, done, allFiles = [], i = 0) {
  fs.readdir(currentPath, function (e, directoryFiles = []) {
    if (!directoryFiles.length)
      return done(allFiles)
    directoryFiles.map(function (file) {
      var joinedPath = path.join(currentPath, file)
      allFiles.push(joinedPath)
      readdir(joinedPath, function () {
        i = i + 1
        if (i == directoryFiles.length)
          done(allFiles)}
      , allFiles)
    })
  })
}

readdir(__dirname, console.log)

注意:两个版本都将跟随符号链接(与原始fs.readdir相同)

There are basically two ways of accomplishing this. In an async environment you'll notice that there are two kinds of loops: serial and parallel. A serial loop waits for one iteration to complete before it moves onto the next iteration - this guarantees that every iteration of the loop completes in order. In a parallel loop, all the iterations are started at the same time, and one may complete before another, however, it is much faster than a serial loop. So in this case, it's probably better to use a parallel loop because it doesn't matter what order the walk completes in, just as long as it completes and returns the results (unless you want them in order).

一个平行循环看起来是这样的:

var fs = require('fs');
var path = require('path');
var walk = function(dir, done) {
  var results = [];
  fs.readdir(dir, function(err, list) {
    if (err) return done(err);
    var pending = list.length;
    if (!pending) return done(null, results);
    list.forEach(function(file) {
      file = path.resolve(dir, file);
      fs.stat(file, function(err, stat) {
        if (stat && stat.isDirectory()) {
          walk(file, function(err, res) {
            results = results.concat(res);
            if (!--pending) done(null, results);
          });
        } else {
          results.push(file);
          if (!--pending) done(null, results);
        }
      });
    });
  });
};

一个串行循环看起来像这样:

var fs = require('fs');
var path = require('path');
var walk = function(dir, done) {
  var results = [];
  fs.readdir(dir, function(err, list) {
    if (err) return done(err);
    var i = 0;
    (function next() {
      var file = list[i++];
      if (!file) return done(null, results);
      file = path.resolve(dir, file);
      fs.stat(file, function(err, stat) {
        if (stat && stat.isDirectory()) {
          walk(file, function(err, res) {
            results = results.concat(res);
            next();
          });
        } else {
          results.push(file);
          next();
        }
      });
    })();
  });
};

并且在你的主目录中测试它(警告:如果你的主目录中有很多东西，结果列表将会非常大):

walk(process.env.HOME, function(err, results) {
  if (err) throw err;
  console.log(results);
});

编辑:改进的示例。

下面是完整的工作代码。按您的要求。您可以递归地获取所有文件和文件夹。

var recur = function(dir) {
            fs.readdir(dir,function(err,list){
                list.forEach(function(file){
                    var file2 = path.resolve(dir, file);
                    fs.stat(file2,function(err,stats){
                        if(stats.isDirectory()) {
                            recur(file2);
                        }
                        else {
                            console.log(file2);
                        }
                    })
                })
            });
        };
        recur(path);

在路径中给出你想要搜索的目录路径，如"c:\test"

下面是一个获得所有文件包括子目录的递归方法。

const FileSystem = require("fs");
const Path = require("path");

//...

function getFiles(directory) {
    directory = Path.normalize(directory);
    let files = FileSystem.readdirSync(directory).map((file) => directory + Path.sep + file);

    files.forEach((file, index) => {
        if (FileSystem.statSync(file).isDirectory()) {
            Array.prototype.splice.apply(files, [index, 1].concat(getFiles(file)));
        }
    });

    return files;
}

因为每个人都应该写自己的，所以我写了一个。

步行(dir, cb, endCb) cb(文件) 零endCb (err |)

脏

module.exports = walk;

function walk(dir, cb, endCb) {
  var fs = require('fs');
  var path = require('path');

  fs.readdir(dir, function(err, files) {
    if (err) {
      return endCb(err);
    }

    var pending = files.length;
    if (pending === 0) {
      endCb(null);
    }
    files.forEach(function(file) {
      fs.stat(path.join(dir, file), function(err, stats) {
        if (err) {
          return endCb(err)
        }

        if (stats.isDirectory()) {
          walk(path.join(dir, file), cb, function() {
            pending--;
            if (pending === 0) {
              endCb(null);
            }
          });
        } else {
          cb(path.join(dir, file));
          pending--;
          if (pending === 0) {
            endCb(null);
          }
        }
      })
    });

  });
}

使用承诺(Q)以函数式风格解决此问题:

var fs = require('fs'),
    fsPath = require('path'),
    Q = require('q');

var walk = function (dir) {
  return Q.ninvoke(fs, 'readdir', dir).then(function (files) {

    return Q.all(files.map(function (file) {

      file = fsPath.join(dir, file);
      return Q.ninvoke(fs, 'lstat', file).then(function (stat) {

        if (stat.isDirectory()) {
          return walk(file);
        } else {
          return [file];
        }
      });
    }));
  }).then(function (files) {
    return files.reduce(function (pre, cur) {
      return pre.concat(cur);
    });
  });
};

它返回一个数组的promise，所以你可以这样使用它:

walk('/home/mypath').then(function (files) { console.log(files); });